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Problem 849

AMC 12 late, AIME early

Geometry Difficulty 4.5 Prove it Euler Olympiad · Russia

In a convex quadrilateral $ABCD$ , the relations $AB = BD$ , $\angle ABD = \angle DBC$ are satisfied. A point $K$ is chosen on diagonal $BD$ so that $BK = BC$ . Prove that $\angle KAD = \angle KCD$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let us mark on side $AB$ a segment $BE = BC$ . The isosceles triangles $EBK$ and $KBC$ are congruent by two sides and the angle between them. Therefore, $EK = KC$ , and $\angle AEK = 180^\circ - \angle BEK = 180^\circ - \angle BKC = \angle CKD$ . Moreover, $KD = BD - BK = BA - BE = EA$ . Hence, triangles $AEK$ and $DKC$ are congruent, from which $\angle KCD = \angle EKA$ .

Further, since both triangles $BEK$ and $BAD$ are isosceles, $\angle BEK = 90^\circ - \angle EBD/2 = \angle BAD$ . Therefore, $AD \parallel EK$ , from which $\angle KAD = \angle EKA = \angle KCD$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.